3.218 \(\int e^{-3 \coth ^{-1}(a x)} (c-a c x)^2 \, dx\)

Optimal. Leaf size=129 \[ -\frac {5}{2} a c^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {35}{3} c^2 x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {35 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}+\frac {1}{3} a^2 c^2 x^3 \sqrt {1-\frac {1}{a^2 x^2}} \]

[Out]

-35/2*c^2*arctanh((1-1/a^2/x^2)^(1/2))/a+16*c^2*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+35/3*c^2*x*(1-1/a^2/x^2)^(1/2)
-5/2*a*c^2*x^2*(1-1/a^2/x^2)^(1/2)+1/3*a^2*c^2*x^3*(1-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6175, 6178, 1805, 1807, 807, 266, 63, 208} \[ \frac {1}{3} a^2 c^2 x^3 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {5}{2} a c^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {35}{3} c^2 x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {35 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^2/E^(3*ArcCoth[a*x]),x]

[Out]

(16*c^2*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + (35*c^2*Sqrt[1 - 1/(a^2*x^2)]*x)/3 - (5*a*c^2*Sqrt[1 - 1/(
a^2*x^2)]*x^2)/2 + (a^2*c^2*Sqrt[1 - 1/(a^2*x^2)]*x^3)/3 - (35*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(2*a)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6175

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*x^p*(1 + c/(d*
x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[n/2] && Inte
gerQ[p]

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int e^{-3 \coth ^{-1}(a x)} (c-a c x)^2 \, dx &=\left (a^2 c^2\right ) \int e^{-3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^2 x^2 \, dx\\ &=-\left (\left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^5}{x^4 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\right )\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {-1+\frac {5 x}{a}-\frac {11 x^2}{a^2}+\frac {15 x^3}{a^3}}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{3} \left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {-\frac {15}{a}+\frac {35 x}{a^2}-\frac {45 x^2}{a^3}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {5}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {1}{6} \left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {-\frac {70}{a^2}+\frac {105 x}{a^3}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {35}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {5}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\left (35 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {35}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {5}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\left (35 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a}\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {35}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {5}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{2} \left (35 a c^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {35}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {5}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {35 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 78, normalized size = 0.60 \[ \frac {1}{6} c^2 \left (\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (2 a^3 x^3-13 a^2 x^2+55 a x+166\right )}{a x+1}-\frac {105 \log \left (a x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )}{a}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a*c*x)^2/E^(3*ArcCoth[a*x]),x]

[Out]

(c^2*((Sqrt[1 - 1/(a^2*x^2)]*x*(166 + 55*a*x - 13*a^2*x^2 + 2*a^3*x^3))/(1 + a*x) - (105*Log[a*(1 + Sqrt[1 - 1
/(a^2*x^2)])*x])/a))/6

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fricas [A]  time = 0.57, size = 104, normalized size = 0.81 \[ -\frac {105 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 105 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (2 \, a^{3} c^{2} x^{3} - 13 \, a^{2} c^{2} x^{2} + 55 \, a c^{2} x + 166 \, c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^2*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

-1/6*(105*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 105*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (2*a^3*c^2*x^3
- 13*a^2*c^2*x^2 + 55*a*c^2*x + 166*c^2)*sqrt((a*x - 1)/(a*x + 1)))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^2*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.06, size = 474, normalized size = 3.67 \[ \frac {\left (2 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} x^{2} a^{2}-15 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{3} a^{3}+4 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} x a +120 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}-30 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{2} a^{2}+15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}-120 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}-46 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+240 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x a -15 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a +30 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}-240 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}+120 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}+15 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -120 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right ) c^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{6 a \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^2*((a*x-1)/(a*x+1))^(3/2),x)

[Out]

1/6*(2*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x^2*a^2-15*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3+4*(a^2)^(1/2)*((a*
x-1)*(a*x+1))^(3/2)*x*a+120*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x^2*a^2-30*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a
^2+15*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3-120*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2
)^(1/2))/(a^2)^(1/2))*x^2*a^3-46*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)+240*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x
*a-15*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x*a+30*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-240*ln(
(a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2+120*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)+15*ln((
a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a-120*a*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)
^(1/2)))/a*c^2*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/((a*x-1)*(a*x+1))^(1/2)/(a*x-1)

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maxima [A]  time = 0.31, size = 204, normalized size = 1.58 \[ -\frac {1}{6} \, a {\left (\frac {105 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {105 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {96 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}} + \frac {2 \, {\left (87 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 136 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 57 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)^2*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

-1/6*a*(105*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 105*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - 96*c
^2*sqrt((a*x - 1)/(a*x + 1))/a^2 + 2*(87*c^2*((a*x - 1)/(a*x + 1))^(5/2) - 136*c^2*((a*x - 1)/(a*x + 1))^(3/2)
+ 57*c^2*sqrt((a*x - 1)/(a*x + 1)))/(3*(a*x - 1)*a^2/(a*x + 1) - 3*(a*x - 1)^2*a^2/(a*x + 1)^2 + (a*x - 1)^3*
a^2/(a*x + 1)^3 - a^2))

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mupad [B]  time = 1.19, size = 163, normalized size = 1.26 \[ \frac {19\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {136\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+29\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{a-\frac {3\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {3\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}}+\frac {16\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}-\frac {35\,c^2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - a*c*x)^2*((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

(19*c^2*((a*x - 1)/(a*x + 1))^(1/2) - (136*c^2*((a*x - 1)/(a*x + 1))^(3/2))/3 + 29*c^2*((a*x - 1)/(a*x + 1))^(
5/2))/(a - (3*a*(a*x - 1))/(a*x + 1) + (3*a*(a*x - 1)^2)/(a*x + 1)^2 - (a*(a*x - 1)^3)/(a*x + 1)^3) + (16*c^2*
((a*x - 1)/(a*x + 1))^(1/2))/a - (35*c^2*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} \left (\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {3 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx + \int \left (- \frac {3 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)**2*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

c**2*(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1), x) + Integral(3*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x +
1))/(a*x + 1), x) + Integral(-3*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1), x) + Integral(a**3*x**
3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1), x))

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